$h(x) = -5x$ $f(n) = 4n^{2}-6n-5(h(n))$ $g(t) = 5t-6+2(h(t))$ $ g(f(-4)) = {?} $
First, let's solve for the value of the inner function, $f(-4)$ . Then we'll know what to plug into the outer function. $f(-4) = 4(-4)^{2}+(-6)(-4)-5(h(-4))$ To solve for the value of $f$ , we need to solve for the value of $h(-4)$ $h(-4) = (-5)(-4)$ $h(-4) = 20$ That means $f(-4) = 4(-4)^{2}+(-6)(-4)+(-5)(20)$ $f(-4) = -12$ Now we know that $f(-4) = -12$ . Let's solve for $g(f(-4))$ , which is $g(-12)$ $g(-12) = (5)(-12)-6+2(h(-12))$ To solve for the value of $g$ , we need to solve for the value of $h(-12)$ $h(-12) = (-5)(-12)$ $h(-12) = 60$ That means $g(-12) = (5)(-12)-6+(2)(60)$ $g(-12) = 54$